\(\int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1264]

Optimal result

Integrand size = 25, antiderivative size = 186 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {(i a-b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}+\frac {2 (b c-a d)}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

-(I*a+b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f+(I* 
a-b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f+2/3*(-a 
*d+b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)-2*(2*a*c*d-b*(c^2-d^2))/(c^2+d^ 
2)^2/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.13 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.62 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \left (-\frac {(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{c-i d}+\frac {(a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{c+i d}\right )}{3 f (c+d \tan (e+f x))^{3/2}} \] Input:

Integrate[(a + b*Tan[e + f*x])/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

((-1/3*I)*(-(((a - I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f* 
x])/(c - I*d)])/(c - I*d)) + ((a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, ( 
c + d*Tan[e + f*x])/(c + I*d)])/(c + I*d)))/(f*(c + d*Tan[e + f*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.97 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4012, 3042, 4012, 3042, 4022, 3042, 4020, 25, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\int \frac {2 b c d+a \left (c^2-d^2\right )-\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 b c d+a \left (c^2-d^2\right )-\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a-i b) (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c-i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a+i b) (c-i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b) (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}}{c^2+d^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a-i b) (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}}{c^2+d^2}\)

Input:

Int[(a + b*Tan[e + f*x])/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(2*(b*c - a*d))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + ((((a - I*b 
)*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + ((a 
+ I*b)*(c - I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/ 
(c^2 + d^2) - (2*(2*a*c*d - b*(c^2 - d^2)))/((c^2 + d^2)*f*Sqrt[c + d*Tan[ 
e + f*x]]))/(c^2 + d^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4012

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4478\) vs. \(2(162)=324\).

Time = 0.44 (sec) , antiderivative size = 4479, normalized size of antiderivative = 24.08

method	result	size
parts	\(\text {Expression too large to display}\)	\(4479\)
derivativedivides	\(\text {Expression too large to display}\)	\(12823\)
default	\(\text {Expression too large to display}\)	\(12823\)

Input:

int((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

a*(-4/f*d*c/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)-1/4/f*d^3/(c^2+d^2)^3*ln(d* 
tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2 
)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+1/f*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2 
)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c 
)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d^5/(c^2+d^2)^(7/2)/(2*(c^2+d^ 
2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2* 
c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-1/f*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d 
^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+ 
e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+2/f*d^5/(c^2+d^2)^(7/2)/(2*(c^2+ 
d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x 
+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/4/f*d^3/(c^2+d^2)^3*ln((c+d*t 
an(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1 
/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2/3/f*d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2 
)-2/f*d/(c^2+d^2)^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e 
))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3 
-2/f*d^3/(c^2+d^2)^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+ 
e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c- 
1/4/f/d/(c^2+d^2)^(7/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d 
^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^5+3/ 
4/f*d^3/(c^2+d^2)^(7/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7181 vs. \(2 (157) = 314\).

Time = 2.08 (sec) , antiderivative size = 7181, normalized size of antiderivative = 38.61 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))**(5/2),x)
 

Output:

Integral((a + b*tan(e + f*x))/(c + d*tan(e + f*x))**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(d-c>0)', see `assume?` for more 
details)Is
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[6,14,5]%%%}+%%%{6,[6,12,5]%%%}+%%%{15,[6,10,5]%%%}+ 
%%%{20,[6
 

Mupad [B] (verification not implemented)

Time = 20.14 (sec) , antiderivative size = 9459, normalized size of antiderivative = 50.85 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(e + f*x))/(c + d*tan(e + f*x))^(5/2),x)
 

Output:

(log(8*b^3*d^16*f^2 - ((((320*b^4*c^2*d^8*f^4 - 16*b^4*d^10*f^4 - 1760*b^4 
*c^4*d^6*f^4 + 1600*b^4*c^6*d^4*f^4 - 400*b^4*c^8*d^2*f^4)^(1/2) + 4*b^2*c 
^5*f^2 + 20*b^2*c*d^4*f^2 - 40*b^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c 
^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*((((( 
320*b^4*c^2*d^8*f^4 - 16*b^4*d^10*f^4 - 1760*b^4*c^4*d^6*f^4 + 1600*b^4*c^ 
6*d^4*f^4 - 400*b^4*c^8*d^2*f^4)^(1/2) + 4*b^2*c^5*f^2 + 20*b^2*c*d^4*f^2 
- 40*b^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^ 
4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(((((320*b^4*c^2*d^8*f^4 - 16*b 
^4*d^10*f^4 - 1760*b^4*c^4*d^6*f^4 + 1600*b^4*c^6*d^4*f^4 - 400*b^4*c^8*d^ 
2*f^4)^(1/2) + 4*b^2*c^5*f^2 + 20*b^2*c*d^4*f^2 - 40*b^2*c^3*d^2*f^2)/(c^1 
0*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8 
*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^22*f^5 + 640*c^3*d^20* 
f^5 + 2880*c^5*d^18*f^5 + 7680*c^7*d^16*f^5 + 13440*c^9*d^14*f^5 + 16128*c 
^11*d^12*f^5 + 13440*c^13*d^10*f^5 + 7680*c^15*d^8*f^5 + 2880*c^17*d^6*f^5 
 + 640*c^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 + 736*b*c^3*d^18*f^4 + 2432*b*c^ 
5*d^16*f^4 + 4480*b*c^7*d^14*f^4 + 4928*b*c^9*d^12*f^4 + 3136*b*c^11*d^10* 
f^4 + 896*b*c^13*d^8*f^4 - 128*b*c^15*d^6*f^4 - 160*b*c^17*d^4*f^4 - 32*b* 
c^19*d^2*f^4 + 96*b*c*d^20*f^4))/4 + (c + d*tan(e + f*x))^(1/2)*(320*b^2*c 
^4*d^14*f^3 - 16*b^2*d^18*f^3 + 1024*b^2*c^6*d^12*f^3 + 1440*b^2*c^8*d^10* 
f^3 + 1024*b^2*c^10*d^8*f^3 + 320*b^2*c^12*d^6*f^3 - 16*b^2*c^16*d^2*f^...
 

Reduce [F]

\[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {-2 \sqrt {d \tan \left (f x +e \right )+c}\, a -3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right )^{2} a \,d^{3} f -6 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right ) a c \,d^{2} f -3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) a \,c^{2} d f +3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right )^{2} b \,d^{3} f +6 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right ) b c \,d^{2} f +3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) b \,c^{2} d f}{3 d f \left (\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}\right )} \] Input:

int((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x)
 

Output:

( - 2*sqrt(tan(e + f*x)*d + c)*a - 3*int((sqrt(tan(e + f*x)*d + c)*tan(e + 
 f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x 
)*c**2*d + c**3),x)*tan(e + f*x)**2*a*d**3*f - 6*int((sqrt(tan(e + f*x)*d 
+ c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3 
*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*a*c*d**2*f - 3*int((sqrt(tan( 
e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2 
*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*a*c**2*d*f + 3*int((sqrt(tan(e 
+ f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d* 
*2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)**2*b*d**3*f + 6*int((sq 
rt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x 
)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*b*c*d**2*f + 3 
*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan 
(e + f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*b*c**2*d*f)/(3*d*f* 
(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2))